From https://en.wikipedia.org/wiki/Skew-symmetric_matrix#Vector_space_structure

The real $n\times n$ matrix ${\textstyle A}$ is skew-symmetric if and only if

${\displaystyle \langle Ax,y\rangle =-\langle x,Ay\rangle \quad \forall x,y\in \mathbb {R} ^{n}}$ This is also equivalent to ${\textstyle \langle x,Ax\rangle =0}$ for all ${\displaystyle x\in \mathbb {R} ^{n}}$

I am trying to show $\langle x,Ax\rangle =0$ but I don't really see it. If I had something like $$\langle x,Ax\rangle = -\langle x,A x\rangle$$ then I can claim $\langle x,Ax\rangle = 0$, but instead I have $$\langle x,Ax\rangle = -\langle x,A^\top x\rangle = -\langle Ax, x\rangle \neq 0 \quad (?)$$


Note that $$ \langle x,Ax\rangle\stackrel{\text{skew symmetry}}{=}-\langle Ax,x\rangle= -\langle x,Ax\rangle\\ \implies 2\langle x,Ax\rangle=0\\ \implies \langle x,Ax\rangle=0 $$

  • $\begingroup$ Oh, essentially the symmetry of the inner product...I see thanks $\endgroup$ – Roy Ayers May 8 at 20:45
  • $\begingroup$ Yes exactly. No problem $\endgroup$ – qbert May 8 at 20:51

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